Question: An isosceles right triangle is removed from each corner of a square piece of paper, as shown, to create a rectangle. If $AB = 12$ units, what is the combined area of the four removed triangles, in square units? [asy]
unitsize(5mm);
defaultpen(linewidth(.7pt)+fontsize(8pt));

pair A=(1,4), Ap=(0,3), B=(3,0), Bp=(4,1);

draw((0,0)--(0,4)--(4,4)--(4,0)--cycle);
draw(A--Ap--B--Bp--cycle,linetype("4 3"));

label("$A$",A,N);
label("$B$",B,S);
[/asy]
Solution: Each of the sides of the square is divided into two segments by a vertex of the rectangle.  Call the lengths of these two segments $r$ and $s$.  Also, let $C$ be the foot of the perpendicular dropped from $A$ to the side containing the point $B$.  Since $AC=r+s$ and $BC=|r-s|$, \[
(r+s)^2+(r-s)^2=12^2,
\] from the Pythagorean theorem.  This simplifies to $2r^2+2s^2=144$, since the terms $2rs$ and $-2rs$ sum to 0.  The combined area of the four removed triangles is $\frac{1}{2}r^2+\frac{1}{2}s^2+\frac{1}{2}r^2+\frac{1}{2}s^2=r^2+s^2$.  From the equation $2r^2+2s^2=144$, this area is $144/2=\boxed{72}$ square units. [asy]
unitsize(5mm);
real eps = 0.4;
defaultpen(linewidth(.7pt)+fontsize(10pt));
pair A=(1,4), Ap=(0,3), B=(3,0), Bp=(4,1);
draw((0,0)--(0,4)--(4,4)--(4,0)--cycle);
draw(A--Ap--B--Bp--cycle,linetype("4 3"));
draw(A--(1,0));
draw(A--B);
draw((1,eps)--(1+eps,eps)--(1+eps,0));
label("$A$",A,N);
label("$B$",B,S);
label("$r$",(4,2.5),E);
label("$s$",(4,0.5),E);
label("$C$",(1,0),S);[/asy]